Question

r1 =7 kω vcc = 12 v r2 =7 kω rc = 4.7 kω β = 15 re = 6.3 kω

Explanation:

To solve for the operating point (e.g., \( I_B \), \( I_C \), \( I_E \), \( V_{CE} \)) of this BJT (Bipolar Junction Transistor) circuit, we follow these steps:

Step 1: Calculate the Base Voltage (\( V_B \))

The base voltage is determined by the voltage divider formed by \( R_1 \) and \( R_2 \).
The formula for \( V_B \) is:

$$ V_B = V_{CC} \cdot \frac{R_2}{R_1 + R_2} $$

Substitute \( V_{CC} = 12 \, \text{V} \), \( R_1 = 7 \, \text{k}\Omega \), and \( R_2 = 7 \, \text{k}\Omega \):

$$ V_B = 12 \cdot \frac{7}{7 + 7} = 12 \cdot \frac{7}{14} = 6 \, \text{V} $$

Step 2: Calculate the Emitter Voltage (\( V_E \))

Assuming the base-emitter voltage (\( V_{BE} \)) is approximately \( 0.7 \, \text{V} \) (for a silicon transistor), we use:

$$ V_E = V_B - V_{BE} $$

Substitute \( V_B = 6 \, \text{V} \) and \( V_{BE} = 0.7 \, \text{V} \):

$$ V_E = 6 - 0.7 = 5.3 \, \text{V} $$

Step 3: Calculate the Emitter Current (\( I_E \))

The emitter current is given by Ohm’s Law through \( R_E \):

$$ I_E = \frac{V_E}{R_E} $$

Substitute \( V_E = 5.3 \, \text{V} \) and \( R_E = 6.3 \, \text{k}\Omega \):

$$ I_E = \frac{5.3}{6.3 \times 10^3} \approx 0.841 \, \text{mA} \, (\text{or } 841 \, \mu\text{A}) $$

Step 4: Calculate the Collector Current (\( I_C \))

For a BJT, \( I_C \approx \beta \cdot I_B \), but for a voltage-divider bias, \( I_C \approx I_E \) (since \( I_E = I_C + I_B \) and \( I_B \ll I_C \)). Alternatively, using \( \beta \):

$$ I_C = \beta \cdot I_B $$

But we can also use \( I_C \approx I_E \) (since \( I_E \approx (1 + \beta) \cdot I_B \), so \( I_C \approx \frac{\beta}{1 + \beta} \cdot I_E \)). For \( \beta = 15 \):

$$ I_C \approx \frac{15}{16} \cdot I_E \approx 0.9375 \cdot 0.841 \approx 0.789 \, \text{mA} $$

(Alternatively, if we assume \( I_C \approx I_E \), \( I_C \approx 0.841 \, \text{mA} \). The difference is small for moderate \( \beta \).)

Step 5: Calculate the Collector Voltage (\( V_C \))

The collector voltage is:

$$ V_C = V_{CC} - I_C \cdot R_C $$

Substitute \( V_{CC} = 12 \, \text{V} \), \( I_C \approx 0.841 \, \text{mA} \), and \( R_C = 4.7 \, \text{k}\Omega \):

$$ V_C = 12 - (0.841 \times 10^{-3}) \cdot (4.7 \times 10^3) \approx 12 - 3.95 \approx 8.05 \, \text{V} $$

Step 6: Calculate the Collector-Emitter Voltage (\( V_{CE} \))

$$ V_{CE} = V_C - V_E $$

Substitute \( V_C \approx 8.05 \, \text{V} \) and \( V_E = 5.3 \, \text{V} \):

$$ V_{CE} \approx 8.05 - 5.3 = 2.75 \, \text{V} $$

Final Values (Summary)

• \( V_B = 6 \, \text{V} \)
• \( V_E = 5.3 \, \text{V} \)
• \( I_E \approx 0.841 \, \text{mA} \)
• \( I_C \approx 0.841 \, \text{mA} \) (or \( 0.789 \, \text{mA} \) using \( \beta \))
• \( V_{CE} \approx 2.75 \, \text{V} \)

(Note: If a specific parameter (e.g., \( I_C \), \( V_{CE} \)) is required, use the corresponding calculation.)

Answer:

To solve for the operating point (e.g., \( I_B \), \( I_C \), \( I_E \), \( V_{CE} \)) of this BJT (Bipolar Junction Transistor) circuit, we follow these steps:

Step 1: Calculate the Base Voltage (\( V_B \))

The base voltage is determined by the voltage divider formed by \( R_1 \) and \( R_2 \).
The formula for \( V_B \) is:

$$ V_B = V_{CC} \cdot \frac{R_2}{R_1 + R_2} $$

Substitute \( V_{CC} = 12 \, \text{V} \), \( R_1 = 7 \, \text{k}\Omega \), and \( R_2 = 7 \, \text{k}\Omega \):

$$ V_B = 12 \cdot \frac{7}{7 + 7} = 12 \cdot \frac{7}{14} = 6 \, \text{V} $$

Step 2: Calculate the Emitter Voltage (\( V_E \))

Assuming the base-emitter voltage (\( V_{BE} \)) is approximately \( 0.7 \, \text{V} \) (for a silicon transistor), we use:

$$ V_E = V_B - V_{BE} $$

Substitute \( V_B = 6 \, \text{V} \) and \( V_{BE} = 0.7 \, \text{V} \):

$$ V_E = 6 - 0.7 = 5.3 \, \text{V} $$

Step 3: Calculate the Emitter Current (\( I_E \))

The emitter current is given by Ohm’s Law through \( R_E \):

$$ I_E = \frac{V_E}{R_E} $$

Substitute \( V_E = 5.3 \, \text{V} \) and \( R_E = 6.3 \, \text{k}\Omega \):

$$ I_E = \frac{5.3}{6.3 \times 10^3} \approx 0.841 \, \text{mA} \, (\text{or } 841 \, \mu\text{A}) $$

Step 4: Calculate the Collector Current (\( I_C \))

For a BJT, \( I_C \approx \beta \cdot I_B \), but for a voltage-divider bias, \( I_C \approx I_E \) (since \( I_E = I_C + I_B \) and \( I_B \ll I_C \)). Alternatively, using \( \beta \):

$$ I_C = \beta \cdot I_B $$

But we can also use \( I_C \approx I_E \) (since \( I_E \approx (1 + \beta) \cdot I_B \), so \( I_C \approx \frac{\beta}{1 + \beta} \cdot I_E \)). For \( \beta = 15 \):

$$ I_C \approx \frac{15}{16} \cdot I_E \approx 0.9375 \cdot 0.841 \approx 0.789 \, \text{mA} $$

(Alternatively, if we assume \( I_C \approx I_E \), \( I_C \approx 0.841 \, \text{mA} \). The difference is small for moderate \( \beta \).)

Step 5: Calculate the Collector Voltage (\( V_C \))

The collector voltage is:

$$ V_C = V_{CC} - I_C \cdot R_C $$

Substitute \( V_{CC} = 12 \, \text{V} \), \( I_C \approx 0.841 \, \text{mA} \), and \( R_C = 4.7 \, \text{k}\Omega \):

$$ V_C = 12 - (0.841 \times 10^{-3}) \cdot (4.7 \times 10^3) \approx 12 - 3.95 \approx 8.05 \, \text{V} $$

Step 6: Calculate the Collector-Emitter Voltage (\( V_{CE} \))

$$ V_{CE} = V_C - V_E $$

Substitute \( V_C \approx 8.05 \, \text{V} \) and \( V_E = 5.3 \, \text{V} \):

$$ V_{CE} \approx 8.05 - 5.3 = 2.75 \, \text{V} $$

Final Values (Summary)

• \( V_B = 6 \, \text{V} \)
• \( V_E = 5.3 \, \text{V} \)
• \( I_E \approx 0.841 \, \text{mA} \)
• \( I_C \approx 0.841 \, \text{mA} \) (or \( 0.789 \, \text{mA} \) using \( \beta \))
• \( V_{CE} \approx 2.75 \, \text{V} \)

(Note: If a specific parameter (e.g., \( I_C \), \( V_{CE} \)) is required, use the corresponding calculation.)